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Reading the Hammack 'Book of Proof'.

He shows that the empty-set is a subset of all possible sets.

I believe this doesn't also mean that the empty-set is also a member of all possible sets.

Is my understanding correct?

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  • $\begingroup$ This is correct. For example, $\varnothing \subset\{1\}$ but $\varnothing\notin\{1\}$. $\endgroup$ – Gyu Eun Lee Nov 15 '16 at 6:01
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Yes, you are correct. Consider the following sets: $A = \{1,2\}$ and $B = \{\emptyset, 1,2\}$. Note that $\emptyset \in B$ and $\emptyset \in B-A = \{\emptyset\}$ but $\emptyset \not\in A$, even though $\emptyset$ is a subset of all the above sets.

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You are right - that does not mean that the emptyset is a member of every set. For example, the emptyset is not a member of the emptyset. (Why? Because the emptyset has no members at all!)

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